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I have noticed one identity in case of tensor product from this post. But I can't understand why it is true.

$\langle v_i| \otimes \langle w_j| \cdot |w_k\rangle \otimes |v_m\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$

Now if I consider $\langle v_i|$ to be $[a,b]$ and $\langle w_j|$ to be $[c,d]$. (note these are $1 \times 2$) then

$\langle v_i| \otimes \langle w_j| = [a,b] \otimes [c,d] = [ac,ad,bc,bd]$

Similarly $|w_k\rangle \otimes |v_m\rangle = [e,f]^T \otimes [g,h]^T = [eg, eh, fg, fh]^T $

Scalar product between them = $[ac,ad,bc,bd] . [eg, eh, fg, fh]^T = aceg+ adeh+ bcfg+ bdfh$

Now as per the identity if we do

$\langle v_i|v_m\rangle = [a,b] . [g,h]^T = ag+bh$ and $\langle w_j|w_k\rangle = [c,d] . [e,f]^T = ce+df$

So $\langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle = (ag+bh).(ce+df)= aceg + adfg + bceh + bdfh$

Now they are not the same. What is the reason for this behavior?

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  • $\begingroup$ Shouldn't ⟨v| = [a,b], ⟨vi| = a, ⟨vm| = b instead of ⟨vi| = [a,b] $\endgroup$ Nov 29 '20 at 7:25
  • $\begingroup$ @AbhayAravinda They do not denote the elements of the entry. Check the answer again. $\endgroup$
    – user27286
    Nov 29 '20 at 8:22
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The identity is wrong. It should be $$ \langle v_i | \otimes \langle w_j | \cdot | w_k \rangle \otimes | v_m \rangle = \langle v_i | w_k \rangle \langle w_j | v_m \rangle. $$

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  • $\begingroup$ If I could give you all the money I would. I can sleep tonight peacefully. Take care and stay well. $\endgroup$
    – user27286
    Nov 29 '20 at 11:08
  • $\begingroup$ @user27286 Note that physicists can sometimes switch the order of tensor products if they know the subsystems from the labels. It's a useful practice, but misleading if you just study this subject. $\endgroup$
    – Danylo Y
    Nov 29 '20 at 12:37
  • $\begingroup$ In this case the more convenient formula is $\langle v_i| \otimes \langle w_j| \cdot |v_m\rangle \otimes |w_k\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$. $\endgroup$
    – Danylo Y
    Nov 29 '20 at 12:42
  • $\begingroup$ @DanyloY Good that you said so...I dont know anything of this subject... Beginner of beginners. $\endgroup$
    – user27286
    Nov 29 '20 at 12:59

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